1. ## doublte integral question

"The double integral $\displaystyle \iint_{R}^{} \, f(x,y) dx\,dy$ can be expressed as $\displaystyle \int_{\pi}^{2\pi}$$\displaystyle \int_{1}^{2} \displaystyle r^2 sin(\theta) dr\,d\Theta in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates? 2. Originally Posted by tennisfreak412 "The double integral \displaystyle \iint_{R}^{} \, dx\,dy can be expressed as \displaystyle \int_{Pi}^{2Pi}$$\displaystyle \int_{1}^{2}$ $\displaystyle r^2 dr\,dTheta$
ummm...what's the question? are you supposed to find the region R?

by the way, type \pi to get $\displaystyle \pi$ and \theta to get $\displaystyle \theta$

3. sorry.. question is here:

"The double integral $\displaystyle \iint_{R}^{} \, f(x,y) dx\,dy$ can be expressed as $\displaystyle \int_{\pi}^{2\pi}$$\displaystyle \int_{1}^{2} \displaystyle r^2 sin(\theta) dr\,d\Theta in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates? 4. Originally Posted by tennisfreak412 "The double integral \displaystyle \iint_{R}^{} \, f(x,y) dx\,dy can be expressed as \displaystyle \int_{\pi}^{2\pi}$$\displaystyle \int_{1}^{2}$ $\displaystyle r^2 sin(\theta) dr\,d\Theta$ in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates?
the radius goes from 1 to 2, the angle theta goes from pi to 2pi. thus the region R is the bottom half of the annulus formed by the circles with common center the origin, and radii 1 and 2 (see the diagram below).

since $\displaystyle \int \int f(x,y)~dxdy \to \int \int f(r \cos \theta, r \sin \theta)~rdrd \theta$ we see that the integral in polar coordinates is $\displaystyle r$. and we know that $\displaystyle r = \sqrt{x^2 + y^2}$

i leave the rest to you

can you continue?

5. I got that, but the difficulty I'm having is doing the double integral for the annulus. I know that I can have the dy limits going from -2 to 0, right? but then how do I do the dx limits so that they incorporate both sides of the hole in the annulus?

6. Originally Posted by tennisfreak412
I got that, but the difficulty I'm having is doing the double integral for the annulus. I know that I can have the dy limits going from -2 to 0, right? but then how do I do the dx limits so that they incorporate both sides of the hole in the annulus?
Hint:

the equation for the inner semi-circle is: $\displaystyle - \sqrt{1 - x^2}$

the equation for the outer semi-circle is: $\displaystyle - \sqrt{4 - x^2}$

can you continue?